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PEN K12

(Canada 1969) Let N = {1, 2, 3, } denote the set of positive integers. Find all

functions f : N N such that for all m, n N: f(2) = 2, f(mn) = f(m)f(n),

f(n + 1) > f(n).

First Solution. To get some idea, we first evaluate f(n) for small positive integers n. It follows

from f(1 1) = f(1) f(1) that f(1) = 1. By the multiplicity, we get f(4) = f(2)2 = 4. It

follows from the inequality 2 = f(2) < f(3) < f(4) = 4 that f(3) = 3. Also, we compute

f(6) = f(2)f(3) = 6. Since 4 = f(4) < f(5) < f(6) = 6, we get f(5) = 5.

We prove by induction that f(n) = n for all n N. We know that it holds for n = 1, 2, 3.

Now, let n > 2 and suppose that f(k) = k for all k {1, , n}. We show that f(n + 1) = n + 1.

Case 1 n + 1 is composite. One may write n + 1 = ab for some positive integers a and b

with 2 a b n. By the inductive hypothesis, we have f(a) = a and f(b) = b. It follows that

f(n + 1) = f(a)f(b) = ab = n + 1.

Case 2 n + 1 is prime. In this case, n + 2 is even. Write n + 2 = 2k for some positive integer

k. Since n 2, we get 2k = n + 2 4 or k 2. Since k = n+22 n, by the inductive hypothesis,

we have f(k) = k. It follows that f(n + 2) = f(2k) = f(2)f(k) = 2k = n + 2. From the inequality

n = f(n) < f(n + 1) < f(n + 2) = n + 2 (1)

we see that f(n + 1) = n + 1.

By induction, we conclude that f(n) = n for all positive integers n.

Second Solution. As in the previous solution, we get f(1) = 1. From the multiplicativity of f, we

find that f(2n) = f(2)f(n) = 2f(n) for all positive integers n. This implies that

f

2k

= 2k (2)

for all positive integers k. Let k N. From the assumption, we obtain the inequality

2

k

= f

2

k< f

2

k

+ 1

< < f

2

k+1

1

< f

2

k+1= 2

k+1

. (3)

In other words, the increasing sequence of 2k + 1 positive integers

f

2k

, f

2k + 1

, , f

2k+1 1

, f

2k+1

(4)

lies in the set of 2k + 1 consecutive integers {2k, 2k + 1, , 2k+1 1, 2k+1}. This means that

f(n) = n for all 2k n 2k+1. Since this holds for all positive integers k, we conclude that

f(n) = n for all n 2.

Third Solution. The assumption that f(mn) = f(m)f(n) for all positive integers m and n is too

strong. We can establish the following proposition.

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Proposition 1. (Putnam 1963/A2) Let f : N N be a strictly increasing function satisfying

that f(2) = 2 and f(mn) = f(m)f(n) for all relatively prime m and n. Then, f is the identity

function onN.

Proof Since f is strictly increasing, we find that f(n + 1) f(n) + 1 for all positive integers

n. It follows that f(n+k) f(n)+k for all positive integers n and k. We now determine p = f(3).

On the one hand, we obtain

f(18) f(15) + 3 f(3)f(5) + 3 f(3)(f(3) + 2) + 3 = p2 + 2p + 3. (5)

On the other hand, we obtain

f(18) = f(2)f(9) 2(f(10) 1) = 2f(2)f(5) 2 4(f(6) 1) 2 = 4f(2)f(3) 6 = 8p 6. (6)

Combining these two, we deduce p2 + 2p + 3 8p 6 or (p 3)2 0. So, we have f(3) = p = 3.

We now prove that f

2l + 1

= 2l + 1 for all positive integers l. Since f(3) = 3, it clearly

holds for l = 1. Assuming that f

2l + 1

= 2l + 1 for some positive integer l, we obtain

f

2l+1 + 2

= f(2)f

2l + 1

= 2

2l + 1

= 2l+1 + 2. (7)

Since f is strictly increasing, this means that f

2l + k

= 2l + k for all k {1, , 2l + 2}. In

particular, we get f

2l+1 + 1

= 2l+1 + 1, as desired.

Now, we find that f(n) = n for all positive integers n. It clearly holds for n = 1, 2. Let l bea fixed positive integer. We have f

2l + 1

= 2l +1 and f

2l+1 + 1

= 2l+1+1. Since f is strictly

increasing, this means that f

2l + k

= 2l + k for all k {1, , 2l + 1}. Since it holds for all

positive integers l, we conclude that f(n) = n for all n 3. This completes the proof.

Fourth Solution. We can establish the following general result.

Proposition 2. Letf : N R+ be a function satisfying the conditions:

(a) f(mn) = f(m)f(n) for all positive integers m and n, and

(b) f(n + 1) f(n) for all positive integers n.

Then, there is a constant R such that f(n) = n for all n N.

Proof We have f(1) = 1. Our job is to show that ln f(n)lnn

is constant when n > 1. Assume to

the contrary thatln f(m)

ln m>

ln f(n)

ln n(8)

for some positive integers m, n > 1. Writing f(m) = mx and f(n) = ny, we have x > y or

ln n

ln m>

ln n

ln m

y

x(9)

So, we can pick a positive rational number AB

, where A, B N, so that

ln n

ln m >

A

B >

ln n

ln m

y

x . (10)

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Hence, mA < nB and mAx > nBy . One the one hand, since f is monotone increasing, the first

inequality mA < nB means that f

mA

f

nB

. On the other hand, since f

mA

= f(m)A

=

mAx

and f

nB

= f(n)B

= nBy

, the second inequality mAx

> nBy

means that

f

mA

= mAx > nBy = f

nB

(11)

This is a contradiction.

Fifth Solution. It is known that we get the same result when we only assume that f is monotone

increasing. In fact, in 1946, Paul Erdos proved the following result in [1]:

Theorem 1. Letf : N R be a function satisfying the conditions:

(a) f(mn) = f(m) + f(n) for all relatively prime m and n, and


Then, there exists a constant R such that f(n) = ln n for all n N.

This implies the following multiplicative result.

Theorem 2. Letf : N R+ be a function satisfying the conditions:

(a) f(mn) = f(m)f(n) for all relatively prime m and n, and


Then, there is a constant R such that f(n) = n for all n N.

Proof1 By Proposition 5, it is enough to show that the function f is completely multi-

plicative: f(mn) = f(m)f(n) for all m and n. We split the proof in three steps.

Step 1 Let a 2 be a positive integer and let a = {x N | gcd(x, a) = 1}. Then, we

obtain

L := infxa

f(x + a)

f(x)= 1 (12)

and

f

ak+1

f

ak

f(a) (13)

for all positive integers k.

Proof of Step 1 Since f is monotone increasing, it is clear that L 1. Now, we notice that

f(k + a) Lf(k) whenever k a. Let m be a positive integer. We take a sufficiently large

integer x0 > ma with gcd (x0, a) = gcd(x0, 2) = 1 to obtain

f(2)f(x0) = f(2x0) f(x0 + ma) Lf(x0 + (m 1)a) Lmf(x0) (14)

or

f(2) Lm. (15)

Since m is arbitrary, this and L 1 force to L = 1. Whenever x a, we obtain

f

ak+1

f(x)

f(ak)=

f

ak+1x

f(ak)

f

ak+1x + ak

f(ak)= f(ax + 1) f

ax + a2

= f(a)f(x + a) (16)

1We present a slightly modified proof in [2]. For another short proof, see [3].

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orf(x + a)

f(x)

f

ak+1

f(a)f(ak). (17)

It follows that 1 = infxaf(x+a)f(x)

f(ak+1)f(a)f(ak)

so that f

ak+1

f

ak

f(a).

Step 2 Similarly, we have

U := supxa

f(x)

f(x + a)= 1 (18)

and

f

ak+1

f

ak

f(a) (19)

for all positive integers k.

Proof of Step 2 The first result immediately follows from Step 1.

supxa

f(x)

f(x + a)=

1

infxaf(x+a)f(x)

= 1. (20)

Whenever x a and x > a, we have

f

ak+1

f(x)

f(ak)=

f

ak+1x

f(ak)

f

ak+1x ak

f(ak)= f(ax 1) f

ax a2

= f(a)f(x a). (21)

It therefore follows that

1 = supxa

f(x)

f(x + a)= sup

xa, x>a

f(x a)

f(x)

f

ak+1

f(a)f(ak). (22)

Step 3 From the two previous results, whenever a 2, we have f

ak+1

= f

ak

f(a).

Then, the straightforward induction gives that

f

ak

= f(a)k

(23)

for all positive integers a and k. Since f is multiplicative, whenever

n = p1k1 pl

kl (24)

gives the standard factorization of n, we obtain

f(n) = fp1

k1

fpl

kl

= f(p1)k1 f(pl)

kl. (25)

We therefore conclude that f is completely multiplicative.

References

1 P. Erdos, On the distribution function of additive functions, Ann. of Math., 47(1946), 1-20

2 E. Howe, A new proof of Erdoss theorem on monotone multiplicative functions, Amer. Math.

Monthly 93(1986), 593-595

3 L. Moser and J. Lambek, On monotone multiplicative functions, Proc. Amer. Math. Soc.,

4(1953), 544-545

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